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AP Physics

Unit 6 - Rotational Motion

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A car is moving up the side of a circular roller coaster loop of radius 12 12 m \text{m} . The angular velocity is 1.8 1.8 rad/s \text{rad/s} and angular acceleration is −0.82 -0.82 rad/s2 \text{rad/s}^2 . The car is at the same elevation as the center of the loop. Find the magnitude and direction (relative to the horizontal) of the acceleration.

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Step Derivation/Formula Reasoning
1 at=α  r=(−0.82 rad/s2)(12 m)=−9.84 m/s2a_{t} = \alpha \; r = (-0.82\,\text{rad/s}^2)(12\,\text{m}) = -9.84\,\text{m/s}^2 This calculates the tangential acceleration by multiplying the angular acceleration α\alpha with the radius rr. The negative sign indicates the tangential acceleration opposes the direction of motion along the track.
2 an=rω2=12 m (1.8 rad/s)2=12 (3.24)=38.88 m/s2a_{n} = r\omega^2 = 12\,\text{m}\,(1.8\,\text{rad/s})^2 = 12\,(3.24) = 38.88\,\text{m/s}^2 This computes the centripetal (normal) acceleration directed toward the center of the circle using the formula an=rω2a_{n} = r\omega^2.
3 a=at2+an2=(−9.84)2+(38.88)2=96.81+1512.83=1609.64≈40.12 m/s2a = \sqrt{a_t^2 + a_n^2} = \sqrt{(-9.84)^2 + (38.88)^2} = \sqrt{96.81 + 1512.83} = \sqrt{1609.64} \approx 40.12\,\text{m/s}^2 The total acceleration is the vector sum of the perpendicular tangential and centripetal components, found using the Pythagorean theorem.
4 θ=arctan⁡(∣at∣an)=arctan⁡(9.8438.88)≈arctan⁡(0.253)≈14.2∘\theta = \arctan\left(\frac{|a_t|}{a_n}\right) = \arctan\left(\frac{9.84}{38.88}\right) \approx \arctan(0.253) \approx 14.2^\circ This angle is measured relative to the horizontal inward (centripetal) direction. Since the tangential acceleration is negative (acting downward while the car moves upward), the net acceleration is directed 14.2∘14.2^\circ below the horizontal inward direction.
5 a≈40.1 m/s2at 14.2∘ below the horizontal (inward)\boxed{a \approx 40.1\,\text{m/s}^2 \quad \text{at } 14.2^\circ \text{ below the horizontal (inward)}} This is the final answer for the magnitude and direction of the acceleration of the car on the loop.

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40.1 m/s214.2∘ below the horizontal (inward)40.1\,\text{m/s}^2 \quad 14.2^\circ\text{ below the horizontal (inward)}

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